escaping the .each { } iteration early in Ruby

Question

code:

 c = 0  
 items.each { |i|  
   puts i.to_s    
   # if c > 9 escape the each iteration early - and do not repeat  
   c++  
 }

I want to grab the first 10 items then leave the "each" loop.

What do I replace the commented line with? is there a better approach? something more Ruby idiomatic?

Answer 1

Does this look like what you want?

10.times { |i|
  puts items[i].to_s
}

Answer 2

There is no ++ operator in Ruby. It's also convention to use do and end for multi-line blocks. Modifying your solution yields:

c = 0  
items.each do |i|  
  puts i.to_s    
  break if c > 9
  c += 1 
end

Or also:

items.each_with_index do |i, c|  
  puts i.to_s    
  break if c > 9
end

See each_with_index and also Programming Ruby Break, Redo, and Next.

Update: Chuck's answer with ranges is more Ruby-like, and nimrodm's answer using take is even better.

Answer 3

items.each_with_index { |i, c| puts i and break if c <= 9 }

Answer 4

break works for escaping early from a loop, but it's more idiomatic just to do items[0..9].each {|i| puts i}. (And if all you're doing is literally printing the items with no changes at all, you can just do puts items[0..9].)

Answer 5

While the break solution works, I think a more functional approach really suits this problem. You want to take the first 10 elements and print them so try

items.take(10).each { |i| puts i.to_s }

Answer 6

Another variant:

puts items.first(10)

Note that this works fine with arrays of less than 10 items:

>> nums = (1..5).to_a
=> [1, 2, 3, 4, 5]
>> puts nums.first(10)
1
2
3
4
5

(One other note, a lot of people are offering some form of puts i.to_s, but in such a case, isn't .to_s redundant? puts will automatically call .to_s on a non-string to print it out, I thought. You would only need .to_s if you wanted to say puts 'A' + i.to_s or the like.)

Answer 7

Another option would be

items.first(10).each do |i|
  puts i.to_s
end

That reads a little more easily to me than breaking on an iterator, and first will return only as many items as available if there aren't enough.